Electric Power - AP Physics C Electricity and Magnetism

A long extension cord is modeled as a $R=0.20\,\Omega$ resistor in series with a load, and the current through the cord is $I=8.0\,\text{A}$ . The cord warms during operation, indicating resistive heating in the wire. Use $P=I^2R$ to determine the power lost as heat in the cord. Give your answer in watts (W). How much power is lost in the transmission line?
Answer: 12.8 W This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes an extension cord with R=0.20 Ω carrying I=8.0 A, requiring calculation of resistive power loss. Choice B (12.8 W) is correct because it applies the formula P=I²R using values I=8.0 A and R=0.20 Ω, resulting in P=(8.0 A)²(0.20 Ω)=64×0.20=12.8 W. Choice D (12.8 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that power loss in wires is continuous dissipation measured in watts. Practice calculating wire losses to understand why thick wires are used for high currents.
A single resistor $R_1$ is connected to an ideal battery, resulting in a total power $P$ delivered by the battery. A second resistor $R_2$ is then connected in parallel with $R_1$ . How does the new total power delivered by the battery compare to $P$ ?
Answer: The new total power is greater than $P$ because the equivalent resistance has decreased. The initial power is $P = V^2/R_1$ . When $R_2$ is added in parallel, the new equivalent resistance is $R_{\text{parallel}} = (1/R_1 + 1/R_2)^{-1}$ . This equivalent resistance is always less than $R_1$ . The new total power is $P_{\text{new}} = V^2/R_{\text{parallel}}$ . Because the denominator (the equivalent resistance) has decreased, the new total power delivered by the battery is greater than the original power $P$ .
To determine which resistor in the circuit dissipates the most power, which of the following provides the sufficient and necessary information?
Answer: The product of the current through and the potential difference across each resistor. The power dissipated by a resistor is given by $P = IV = I^2R = V^2/R$ . Since the resistors are not identical and the currents and voltages are generally different for each, knowing only the current (A), voltage (B), or resistance (C) is insufficient. For example, a large current in a small resistor may dissipate less power than a smaller current in a large resistor. One must calculate the power for each resistor using one of the formulas, which requires knowing $I$ and $V$ , or $I$ and $R$ , or $V$ and $R$ . The product $IV$ is the definition of power and is always sufficient.
A $50 \, \Omega$ resistor is connected to a power source, resulting in a constant current of $0.50 \, \text{A}$ . How much total thermal energy is dissipated in the resistor during a $2.0$ -minute interval?
Answer: $1500 \, \text{J}$ First, calculate the power dissipated: $P = I^2R = (0.50 \, \text{A})^2 (50 \, \Omega) = 12.5 \, \text{W}$ . Next, convert the time interval to seconds: $t = 2.0 \, \text{min} \times 60 \, \text{s/min} = 120 \, \text{s}$ . The total energy dissipated is $E = P \times t = 12.5 \, \text{W} \times 120 \, \text{s} = 1500 \, \text{J}$ .
Two identical light bulbs, each with resistance $R$ , are connected to an ideal battery with voltage $V$ . What is the ratio of the total power consumed when they are in parallel to the total power consumed when they are in series, $P_{\text{parallel}}/P_{\text{series}}$ ?
Answer: $4$ In series, the total resistance is $R_{\text{series}} = R + R = 2R$ . The total power is $P_{\text{series}} = V^2 / R_{\text{series}} = V^2 / (2R)$ . In parallel, the equivalent resistance is $R_{\text{parallel}} = (1/R + 1/R)^{-1} = R/2$ . The total power is $P_{\text{parallel}} = V^2 / R_{\text{parallel}} = V^2 / (R/2) = 2V^2/R$ . The ratio is $\frac{P_{\text{parallel}}}{P_{\text{series}}} = \frac{2V^2/R}{V^2/(2R)} = 4$ .
A battery has an emf $\mathcal{E}$ and an internal resistance $r$ . It is connected in series with an external resistor of resistance $R$ . What is the total power supplied by the source of emf?
Answer: $\frac{\mathcal{E}^2}{R+r}$ The total resistance of the circuit is $R_{\text{total}} = R+r$ . The current flowing in the circuit is $I = \frac{\mathcal{E}}{R+r}$ . The total power supplied by the source of emf is given by $P_{\text{source}} = I\mathcal{E}$ . Substituting the expression for $I$ , we get $P_{\text{source}} = \left(\frac{\mathcal{E}}{R+r}\right)\mathcal{E} = \frac{\mathcal{E}^2}{R+r}$ . This power is dissipated in both the internal and external resistances.
A power supply has an emf of $24 \, \text{V}$ and an internal resistance of $2.0 \, \Omega$ . To which of the following load resistances will the power supply deliver the most power?
Answer: $2.0 \, \Omega$ The maximum power transfer theorem states that a source delivers maximum power to a load when the load resistance equals the internal resistance of the source. In this case, the internal resistance is $2.0 \, \Omega$ , so the maximum power will be delivered to a load resistance of $2.0 \, \Omega$ .
A resistor of resistance $R$ is connected to Battery A, which has emf $\mathcal{E}$ and internal resistance $r_A$ , dissipating power $P_A$ . The same resistor is then connected to Battery B, which also has emf $\mathcal{E}$ but a smaller internal resistance $r_B < r_A$ . The power dissipated in R is now $P_B$ . How does $P_A$ compare to $P_B$ ?
Answer: $P_A < P_B$ The power dissipated in the external resistor is $P = I^2 R = \left(\frac{\mathcal{E}}{R+r}\right)^2 R$ . Since $\mathcal{E}$ and $R$ are the same for both cases, power depends on the total resistance $R+r$ . As $r_B < r_A$ , the total resistance of the circuit with Battery B, $R+r_B$ , is less than that with Battery A, $R+r_A$ . A smaller total resistance results in a larger current. Since power is proportional to the square of the current, $P_B > P_A$ .
A desk lamp uses a bulb modeled as a resistor of $R=144\,\Omega$ connected across a $12.0\,\text{V}$ DC supply. The lamp is used in a low-voltage lighting circuit with negligible wire resistance. Use $P=\frac{V^2}{R}$ to determine the bulb's power usage. Report the answer in watts (W). What is the power usage of the light bulb?
Answer: 1.0 W This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a 144 Ω bulb connected across 12.0 V DC, requiring calculation of power using P=V²/R. Choice A (1.0 W) is correct because it applies the formula P=V²/R using values V=12.0 V and R=144 Ω, resulting in P=(12.0 V)²/(144 Ω)=144/144=1.0 W. Choice D (144 W) is incorrect due to using V² without dividing by R, often a result of incomplete formula application. To help students, emphasize careful formula application and unit checking. Practice with low-voltage DC circuits to build familiarity with typical power values.
An inductor $L$ and a resistor $R$ are connected in series with a battery of emf $\mathcal{E}$ and a switch. The switch is closed at time $t=0$ . What is the power dissipated by the resistor as a function of time $t$ ?
Answer: $\frac{\mathcal{E}^2}{R} (1 - e^{-Rt/L})^2$ The current in a charging LR circuit is $I(t) = \frac{\mathcal{E}}{R}(1 - e^{-Rt/L})$ . The power dissipated in the resistor is $P_R(t) = (I(t))^2 R$ . Substituting the expression for $I(t)$ gives $P_R(t) = \left[\frac{\mathcal{E}}{R}(1 - e^{-Rt/L})\right]^2 R = \frac{\mathcal{E}^2}{R^2}(1 - e^{-Rt/L})^2 R = \frac{\mathcal{E}^2}{R}(1 - e^{-Rt/L})^2$ .
A single resistor $R_1$ is connected to an ideal battery, dissipating a total power $P$ . A second resistor $R_2$ is then connected in series with $R_1$ . How does the new total power dissipated by the circuit compare to $P$ ?
Answer: The new total power is less than $P$ because the total resistance has increased. The initial power is $P = V^2/R_1$ . When $R_2$ is added in series, the new total resistance is $R_{\text{series}} = R_1 + R_2$ . Since $R_1+R_2 > R_1$ , the new total resistance is greater than the original resistance. The new total power is $P_{\text{new}} = V^2/R_{\text{series}}$ . Because the denominator has increased, the new total power is less than the original power $P$ .
A 18.0 V battery powers a portable circuit with two resistors in parallel, $R_1=9.0\,\Omega$ and $R_2=18.0\,\Omega$ , giving total resistance $R_\text{eq}=6.0\,\Omega$ . The total current drawn from the battery is $I=3.0\,\text{A}$ . Use $P=IV$ to compute the power supplied by the battery. Report power in watts (W). Calculate the power supplied by the battery.
Answer: 54 W This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes an 18.0 V battery with total current I=3.0 A, requiring calculation of power supplied by the battery. Choice B (54 W) is correct because it applies the formula P=IV using values I=3.0 A and V=18.0 V, resulting in P=(3.0 A)(18.0 V)=54 W. Choice D (54 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that battery power output is always P=IV where I is total current drawn and V is battery voltage. Practice with parallel circuits to understand how total current relates to individual branch currents.
A power source has a fixed emf $\mathcal{E}$ and a fixed internal resistance $r$ . To which value of external load resistance $R$ will the source deliver the maximum power?
Answer: $R = r$ According to the maximum power transfer theorem, a source with internal resistance $r$ delivers maximum power to an external load resistor $R$ when $R=r$ . This can be proven by taking the derivative of the power expression $P = I^2 R = (\frac{\mathcal{E}}{R+r})^2 R$ with respect to $R$ and setting it to zero.
A 9.0 V battery powers a wearable device modeled as two resistors in series: $R_1=3.0\,\Omega$ and $R_2=6.0\,\Omega$ , so the total resistance is $9.0\,\Omega$ . The same current flows through both resistors and equals $1.0\,\text{A}$ . Determine the power dissipated by $R_2$ using $P=I^2R$ . Give your answer in watts (W). What is the power dissipated by the resistor in the circuit?
Answer: 6.0 W This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes two resistors in series with R₂=6.0 Ω and current I=1.0 A, requiring calculation of power dissipated by R₂. Choice A (6.0 W) is correct because it applies the formula P=I²R using values I=1.0 A and R=6.0 Ω, resulting in P=(1.0 A)²(6.0 Ω)=6.0 W. Choice D (6.0 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize unit consistency and that power dissipation in a resistor is always P=I²R when current is known. Practice recognizing series circuits where the same current flows through all components.
A student analyzes a series circuit with a battery and two different resistors, $R_1$ and $R_2$ . The student claims, "The total power dissipated by the two resistors in series is the sum of the powers each would dissipate if connected individually to the same battery." Which of the following is a correct evaluation of this claim?
Answer: The claim is incorrect, because the current in the series circuit is different from the current through each resistor when connected individually. The claim is incorrect. Let the battery voltage be $V$ . The power dissipated by $R_1$ alone is $P_1 = V^2/R_1$ , and by $R_2$ alone is $P_2 = V^2/R_2$ . Their sum is $V^2(1/R_1 + 1/R_2)$ . In series, the total resistance is $R_s = R_1 + R_2$ , and the total power is $P_s = V^2/(R_1 + R_2)$ . These expressions are not equal. The physical reason is that the current drawn from the battery changes. In the series circuit, the current is $I_s = V/(R_1+R_2)$ , which is smaller than the current through either resistor when connected alone.
In a series RC circuit connected to a battery of emf $\mathcal{E}$ , the switch is closed at $t=0$ . Which statement correctly describes the energy transformation in the circuit during the charging process?
Answer: The instantaneous power delivered by the battery equals the sum of the rate of energy dissipation in the resistor and the rate of energy storage in the capacitor. The power delivered by the battery is $P_{batt} = I(t)\mathcal{E}$ . This power is used for two purposes: dissipating energy as heat in the resistor at a rate $P_R = I(t)^2 R$ , and storing energy in the capacitor's electric field at a rate $P_C = dU_C/dt = V_C(t)I(t)$ . By conservation of energy, $P_{batt} = P_R + P_C$ . Since the current $I(t)$ changes with time, these are all instantaneous rates.
A capacitor of capacitance $C$ is charged to an initial potential difference $V_0$ . At time $t=0$ , it is connected across a resistor $R$ and begins to discharge. What is the total energy that will be dissipated as heat by the resistor during the entire discharging process?
Answer: $\frac{1}{2} C V_0^2$ By the principle of conservation of energy, the total energy dissipated by the resistor must be equal to the total energy initially stored in the capacitor. The initial energy stored in the capacitor is given by $U_C = \frac{1}{2} C V_0^2$ . This entire amount of energy is converted into thermal energy in the resistor as the capacitor discharges to zero potential.
An ideal, constant voltage source $V$ is connected to a variable resistor of resistance $R$ . How does the power $P$ dissipated by the resistor depend on its resistance $R$ ?
Answer: $P$ is inversely proportional to $R$ . For a circuit with a constant voltage source, the most convenient formula for power is $P = V^2/R$ . Since $V$ is constant, this equation shows that the power $P$ is inversely proportional to the resistance $R$ . As $R$ increases, $P$ decreases.
Two resistors, $R_1$ and $R_2$ , with $R_1 > R_2$ , are connected in parallel to an ideal battery. Which resistor dissipates more power?
Answer: $R_2$ , because it has a smaller resistance and the potential difference is the same across both. For components in parallel, the potential difference $V$ is the same across each. The power dissipated by a resistor is given by $P = V^2/R$ . Since $V$ is the same for both resistors and $R_1 > R_2$ , the resistor with the smaller resistance, $R_2$ , will dissipate more power because power is inversely proportional to resistance.
A heating element is rated at $2.0$ kilowatts. For approximately how long must this element operate to dissipate $4.0 \times 10^5$ joules of energy?
Answer: $200 \, \text{s}$ Power is the rate of energy transfer, $P = E/t$ . Therefore, the time required is $t = E/P$ . Given $P = 2.0 \, \text{kW} = 2000 \, \text{W}$ (or J/s) and $E = 4.0 \times 10^5 \, \text{J}$ . So, $t = (4.0 \times 10^5 \, \text{J}) / (2000 \, \text{J/s}) = 200 \, \text{s}$ .

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